Question

VI. A recent report published in USA Today indicated that a common family of four spends $ 490 a month on food. Assume that the distribution of food expenses for a family of four follows a symmetric distribution, with an average dispersion of $ 90. Based on the information obtained, we want to identify: what percentage of families do you spend: more than $ 30 and less than $ 490 on food per month; less than less than $ 430 per month in food; between $ 430 and $ 600 per month in food; between $ 500 and $ 600 per month in food; and less than $ 410 or more than 570 per month in food. Perform the analysis required for the study detailing an individual interpretation sentence for each situation.

Answer 1

Answer:

Given,

mean = 490

s = 90

a)

P(30 < x < 490) = P((30 - 490)/90 < (x-mu)/s <= (490-490)90)

= P(-5.11 < z < 0)

= P(z < 0) - P(z < -5.11)

= 0.5 - 0 [since from z table]

= 0.5

b)

P(x < 430)

= P((x-mu)/s < (430 - 490)/90)

= P(z < -0.67)

= 0.2514 [since from z table]

c)

P(430 < x < 600) = P((430 - 490)/90 < (x-mu)/s < (600 - 490)/90)

= P(-0.67 < z < 1.22)

= P(z < 1.22) - P(z < -0.67)

= 0.8888 - 0.2514

= 0.6374 [since from z table]